Case: Meet in the Middle

$C_1 + C_2 = N+1$

When we have a lane with clues on both sides that add up to $N+1$ , we can pinpoint the lane peak.


3			6				4

Examples

Case 1




2					3

The clues add up to $2 + 3 = 5$ , which is $1$ above $4$ (since this a 4x4 puzzle), so the case applies.




2	4	3

Case 2


2	3	3






5	4	3

This is a 6x6 puzzle, so we require a sum of $6 + 1 = 7$ .

$2 + 5 = 7$ and $3 + 4 = 7$ , so we can pinpoint the lane peaks in those lanes.


2	3	3

6
	6



5	4	3

We can’t apply the case to the remaining lane, but we can actually still pinpoint the $6$ ! Together, the $3$ -clues constrict it to these two cells:


2	3	3

6
	6	6
		6


5	4	3

But of course, in the upper cell’s row the $6$ has already been taken, so we’re left with just the lower cell.


2	3	3

6
	6
		6


5	4	3

Explanation

Take a look at the constraints on where the $6$ can go produced by the $3$ and $4$ clues:


3			6	6	6	6
	6	6	6				4

There’s only 1 spot for the $6$ !

This happens whenever the 2 clues add up to $N+1$ , which is in fact the highest any 2 opposite clues can ever add up to. If they added up to more, the lane would be unsolvable!


4							4
4				6	6	6
	6	6	6				4

Nowhere for $6$ to go!

This deduction is a small reason why closed Skyscrapers are easier than open ones – since every clue has an opposite clue, it’s far more likely you’ll be able to use this deduction to pinpoint a lane peak!